Question

Find the equation of the plane in xyz-space through the point P = (4, 4, 5) and perpendicular to the vector n =( -4, -4, -3.)

Answer 1

The equation of the plane is -4(x-4) - 4(y-4) - 3(z-5) = 0

Step-by-step explanation:

To find the equation of the plane, we need a point on the plane and a vector perpendicular to the plane. The given point is P = (4, 4, 5) and the given vector is n = (-4, -4, -3).

Since the plane is perpendicular to the given vector, the normal vector of the plane will be the same as the given vector. Therefore, the equation of the plane is:

-4(x-4) - 4(y-4) - 3(z-5) = 0